Question: $g(n) = 6n^{2}+f(n)$ $h(t) = -4t^{3}+3t^{2}-5t-4(g(t))$ $f(t) = -4t^{2}-3t-3$ $ f(g(1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $g(1)$ . Then we'll know what to plug into the outer function. $g(1) = 6(1^{2})+f(1)$ To solve for the value of $g$ , we need to solve for the value of $f(1)$ $f(1) = -4(1^{2})+(-3)(1)-3$ $f(1) = -10$ That means $g(1) = 6(1^{2})-10$ $g(1) = -4$ Now we know that $g(1) = -4$ . Let's solve for $f(g(1))$ , which is $f(-4)$ $f(-4) = -4(-4)^{2}+(-3)(-4)-3$ $f(-4) = -55$